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Assoli18 [71]
3 years ago
6

How long will it take to melt 3500 kg of steel in a 50 kW oven operating at 65% efficiency from room temperature to 1600°C?

Physics
1 answer:
zepelin [54]3 years ago
4 0

Answer:

71195 seconds or 19.8 hours

Explanation:

Because the steel will be melted, beginning from room temperature, energy is needed to heat it from room temperature to 1600°C and another to melt it at 1600°C.

We take room temperature to be 27°C, the specific heat capacity of steel to be 420 J/kg/K and the specific latent heat of fusion of steel to be 440 J/kg.

The heat required to heat the steel from room temperature to 1600°C is given by

H_S = mc_S\delta\theta

m is the mass, c_S is the specific heat capacity and \delta\theta is the change in temperature.

H_S = 3500\times420\times(1600-27) = 2312310000

The heat required to melt it at 1600°C is

H_L = ml_S

m is the mass and l_S is the specific latent heat of fusion of steel.

H_L = 3500\times440 = 1540000

Adding both, the heat required is 2312310000+1540000=2313850000\text{ J}

Because the oven is 65% efficient, its output power = 65\%\times50000 = 32500 \text{ W}

Now, energy = power * time

Time = Energy/power

t = \dfrac{2313850000}{32500}=71195 \text{ seconds} = 19.8 \text{ seconds}

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